JEE Mathematics Vector Algebra MCQs Set 19

Answer: (b) \( 6\hat{i}−2\hat{j}+6\hat{k} \)

Explanation:
1. \( \vec{a}×\vec{b}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&1&1\\1&-1&1\end{vmatrix}=2\hat{i}−2\hat{k} \).
2. \( (\vec{a}×\vec{b})×\vec{c}=(2,0,−2)×(2,3,−1)=(0·(−1)−(−2)·3,\ (−2)·2−2·(−1),\ 2·3−0)=(6,−2,6) \).
3. So the result is \( 6\hat{i}−2\hat{j}+6\hat{k} \).

Teacher's Note:
1. Always compute the inner cross product first and keep the order of the factors.
2. Swapping the order of a cross product changes the sign.

Answer: (b) 0

Explanation:
1. \( (\vec{a}×\vec{b})×\vec{c}=(\vec{a}·\vec{c})\vec{b}−(\vec{b}·\vec{c})\vec{a} \).
2. \( \vec{a}·\vec{c}=1 \) and \( \vec{b}·\vec{c}=1 \), so \( (\vec{a}×\vec{b})×\vec{c}=\vec{b}−\vec{a} \).
3. Hence \( λ=−1, μ=1 \) and \( λ+μ=0 \).

Teacher's Note:
1. Use the identity for the vector triple product (middle vector times dot product with the third, minus the first times dot product of the other two).
2. Then compare coefficients of the independent vectors.

Answer: (d) \( \sqrt5 \)

Explanation:
1. \( \vec{a}·\vec{c}=8+6−12=2 \) and \( \vec{a}·\vec{b}=2+3−4=1 \).
2. \( \vec{a}×(\vec{b}×\vec{c})=2\vec{b}−1·\vec{c}=(2−4,\ 2−2,\ 2−3)=(−2,0,−1) \).
3. Its magnitude is \( \sqrt{4+0+1}=\sqrt5 \).

Teacher's Note:
1. Evaluating the two dot products first is much shorter than finding two cross products.
2. Check the final magnitude with the Pythagoras formula.

Answer: (b) The null vector \( \vec0 \)

Explanation:
1. Expand each term: \( (\vec{a}·\vec{c})\vec{b}−(\vec{a}·\vec{b})\vec{c} \), \( (\vec{b}·\vec{a})\vec{c}−(\vec{b}·\vec{c})\vec{a} \) and \( (\vec{c}·\vec{b})\vec{a}−(\vec{c}·\vec{a})\vec{b} \).
2. Adding, every term cancels in pairs.
3. The result is the null vector (the Jacobi identity).

Teacher's Note:
1. The result is a vector, so it is the null vector, not the scalar zero.
2. This identity is used to simplify sums of cyclic triple products.

Answer: (a) \( (\vec{a}×\vec{c})×\vec{b}=\vec0 \)

Explanation:
1. \( (\vec{a}×\vec{b})×\vec{c}=(\vec{a}·\vec{c})\vec{b}−(\vec{b}·\vec{c})\vec{a} \) and \( \vec{a}×(\vec{b}×\vec{c})=(\vec{a}·\vec{c})\vec{b}−(\vec{a}·\vec{b})\vec{c} \).
2. Equating gives \( (\vec{a}·\vec{b})\vec{c}−(\vec{b}·\vec{c})\vec{a}=\vec0 \).
3. But \( (\vec{a}×\vec{c})×\vec{b}=(\vec{a}·\vec{b})\vec{c}−(\vec{c}·\vec{b})\vec{a} \), which is exactly this expression, so the condition is \( (\vec{a}×\vec{c})×\vec{b}=\vec0 \).

Teacher's Note:
1. Expand both sides and compare, then recognise the remaining expression as another triple product.
2. The vector triple product is not associative in general.

Answer: (c) Both \( \vec{c} \) and \( \vec{a}×\vec{b} \)

Explanation:
1. A cross product is perpendicular to both of its factors.
2. The factors of \( (\vec{a}×\vec{b})×\vec{c} \) are \( \vec{a}×\vec{b} \) and \( \vec{c} \).
3. So it is perpendicular to both of them.

Teacher's Note:
1. Read the outermost cross product to see which two vectors the result is perpendicular to.
2. The result lies in the plane of \( \vec{a} \) and \( \vec{b} \).

Answer: (b) \( 2\vec{a} \)

Explanation:
1. \( \hat{i}×(\vec{a}×\hat{i})=(\hat{i}·\hat{i})\vec{a}−(\hat{i}·\vec{a})\hat{i}=\vec{a}−a_1\hat{i} \).
2. Sum over the three unit vectors: \( 3\vec{a}−(a_1\hat{i}+a_2\hat{j}+a_3\hat{k})=3\vec{a}−\vec{a} \).
3. The sum equals \( 2\vec{a} \).

Teacher's Note:
1. This is a standard result: summing over the three axes gives twice the vector.
2. Expand each term with the triple product formula.

Answer: (d) 4

Explanation:
1. Use \( (\vec{a}×\vec{b})·(\vec{c}×\vec{d})=(\vec{a}·\vec{c})(\vec{b}·\vec{d})−(\vec{a}·\vec{d})(\vec{b}·\vec{c}) \).
2. \( \vec{a}·\vec{c}=4, \vec{b}·\vec{d}=−1+(−2)+4=1, \vec{a}·\vec{d}=2−3+1=0 \).
3. So the value is \( 4×1−0=4 \).

Teacher's Note:
1. The scalar product of two cross products is a 2 × 2 determinant of dot products.
2. Compute the four dot products carefully.

Answer: (b) \( |\vec{b}|^2 \)

Explanation:
1. \( (\vec{a}×\vec{b})·(\vec{c}×\vec{b})=(\vec{a}·\vec{c})(\vec{b}·\vec{b})−(\vec{a}·\vec{b})(\vec{b}·\vec{c}) \).
2. Adding \( (\vec{a}·\vec{b})(\vec{b}·\vec{c}) \) gives \( |\vec{b}|^2(\vec{a}·\vec{c}) \).
3. Comparing with \( λ(\vec{a}·\vec{c}) \) gives \( λ=|\vec{b}|^2 \).

Teacher's Note:
1. Apply the Lagrange-type identity for the dot product of two cross products.
2. The terms with \( \vec{a}·\vec{b} \) cancel.

Answer: (d) \( 2(\vec{a}·\vec{b}) \)

Explanation:
1. \( (\vec{a}×\hat{i})·(\vec{b}×\hat{i})=(\vec{a}·\vec{b})(\hat{i}·\hat{i})−(\vec{a}·\hat{i})(\hat{i}·\vec{b})=\vec{a}·\vec{b}−a_1b_1 \).
2. Adding the three terms: \( 3(\vec{a}·\vec{b})−(a_1b_1+a_2b_2+a_3b_3) \).
3. This equals \( 2(\vec{a}·\vec{b}) \).

Teacher's Note:
1. Use the same four-vector identity for each axis vector.
2. The leftover sum of products is the dot product itself.

Answer: (a) \( |\vec{a}|^2(\vec{b}·\vec{c}) \)

Explanation:
1. \( (\vec{a}×\vec{b})·(\vec{a}×\vec{c})=|\vec{a}|^2(\vec{b}·\vec{c})−(\vec{a}·\vec{b})(\vec{a}·\vec{c}) \).
2. Since \( \vec{a} \) is perpendicular to both \( \vec{b} \) and \( \vec{c} \) (being parallel to \( \vec{b}×\vec{c} \)), \( \vec{a}·\vec{b}=\vec{a}·\vec{c}=0 \).
3. Hence the value is \( |\vec{a}|^2(\vec{b}·\vec{c}) \).

Teacher's Note:
1. A vector parallel to \( \vec{b}×\vec{c} \) is perpendicular to both \( \vec{b} \) and \( \vec{c} \).
2. That kills the second term of the identity.

Answer: (c) 4

Explanation:
1. \( [\vec{a}\ \vec{b}\ \vec{a}×\vec{b}]=(\vec{a}×\vec{b})·(\vec{a}×\vec{b}) \) (the triple product is cyclic).
2. This equals \( |\vec{a}×\vec{b}|^2=4 \).

Teacher's Note:
1. Rotate the scalar triple product so that the cross product comes first.
2. A vector dotted with itself gives the square of its magnitude.

Answer: (a) \( 24(\hat{i}+\hat{j}−2\hat{k}) \)

Explanation:
1. \( \vec{a}×\vec{b}=(1−6,\ 9−2,\ 4−3)=(−5,7,1) \) and \( \vec{c}×\vec{d}=(1+8,\ −4+1,\ 2+1)=(9,−3,3) \).
2. \( (−5,7,1)×(9,−3,3)=(21+3,\ 9+15,\ 15−63)=(24,24,−48) \).
3. This equals \( 24(\hat{i}+\hat{j}−2\hat{k}) \).

Teacher's Note:
1. Find the two cross products separately and then cross them.
2. Factor out the common number to match the option.

Answer: (d) The null vector

Explanation:
1. For coplanar vectors, \( \vec{a}×\vec{b} \) and \( \vec{c}×\vec{d} \) are both perpendicular to the same plane.
2. So they are parallel and their cross product is zero.
3. The answer is the null vector.

Teacher's Note:
1. Two cross products of vectors lying in one plane are both normal to that plane.
2. Parallel vectors have a zero cross product.

Answer: (c) 9

Explanation:
1. \( (\vec{b}×\vec{c})×(\vec{c}×\vec{a})=[\vec{a}\ \vec{b}\ \vec{c}]\vec{c} \), so \( [\vec{a}\ \vec{b}\ \vec{c}]=3 \).
2. \( [\vec{b}×\vec{c}\ \vec{c}×\vec{a}\ \vec{a}×\vec{b}]=[\vec{a}\ \vec{b}\ \vec{c}]^2 \).
3. The value is \( 3^2=9 \).

Teacher's Note:
1. Memorise: the scalar triple product of the three cross products is the square of \( [\vec{a}\ \vec{b}\ \vec{c}] \).
2. The given condition gives \( [\vec{a}\ \vec{b}\ \vec{c}] \) directly.

Answer: (b) \( 2\hat{i}−2\hat{j} \)

Explanation:
1. \( \vec{a}·\vec{c}=1−1+1=1 \) and \( \vec{a}·\vec{b}=1−1−1=−1 \).
2. \( \vec{a}×(\vec{b}×\vec{c})=(\vec{a}·\vec{c})\vec{b}−(\vec{a}·\vec{b})\vec{c}=\vec{b}+\vec{c} \).
3. \( \vec{b}+\vec{c}=2\hat{i}−2\hat{j} \).

Teacher's Note:
1. Use the expansion \( \vec{a}×(\vec{b}×\vec{c})=(\vec{a}·\vec{c})\vec{b}−(\vec{a}·\vec{b})\vec{c} \) and substitute the components.
2. Take care with the signs of the dot products.

Answer: (a) −4

Explanation:
1. \( \vec{a}·\vec{c}=1−6+6=1 \) and \( \vec{a}·\vec{b}=2−2+3=3 \).
2. \( \vec{a}×(\vec{b}×\vec{c})=\vec{b}−3\vec{c}=(2−3,\ 1−9,\ −1+6)=(−1,−8,5) \).
3. \( p+q+r=−1−8+5=−4 \).

Teacher's Note:
1. Use the expansion \( \vec{a}×(\vec{b}×\vec{c})=(\vec{a}·\vec{c})\vec{b}−(\vec{a}·\vec{b})\vec{c} \) and substitute the components.
2. Add the components only at the very end.

Answer: (c) 0

Explanation:
1. \( \vec{p}·\vec{r}=4−10+6=0 \) and \( \vec{p}·\vec{q}=6−10+4=0 \).
2. \( \vec{p}×(\vec{q}×\vec{r})=(\vec{p}·\vec{r})\vec{q}−(\vec{p}·\vec{q})\vec{r}=\vec0 \).
3. So the magnitude is 0.

Teacher's Note:
1. Check the dot products before computing any cross product; here both vanish.
2. A quick check saves a long calculation.

Answer: (d) The null vector

Explanation:
1. Here \( \hat{j}×\hat{k}=\hat{i} \), so \( \hat{i}×\hat{i}=\vec0 \).
2. Similarly \( \hat{k}×\hat{i}=\hat{j} \) gives \( \hat{j}×\hat{j}=\vec0 \) and \( \hat{i}×\hat{j}=\hat{k} \) gives \( \hat{k}×\hat{k}=\vec0 \).
3. The sum is the null vector.

Teacher's Note:
1. Evaluate the brackets first using the cyclic rule \( \hat{i}×\hat{j}=\hat{k} \).
2. Each term becomes the cross product of a vector with itself.

Answer: (b) Parallel

Explanation:
1. Expanding: \( (\vec{a}·\vec{c})\vec{b}−(\vec{b}·\vec{c})\vec{a}=(\vec{a}·\vec{c})\vec{b}−(\vec{a}·\vec{b})\vec{c} \).
2. So \( (\vec{a}·\vec{b})\vec{c}=(\vec{b}·\vec{c})\vec{a} \).
3. Since both scalars are non-zero, \( \vec{c} \) is a scalar multiple of \( \vec{a} \), i.e. they are parallel.

Teacher's Note:
1. Cancel the common term and read off the proportionality.
2. The conditions \( \vec{a}·\vec{b}≠0 \), \( \vec{b}·\vec{c}≠0 \) ensure the multiples are non-zero.

Answer: (c) Coplanar with \( \vec{b} \) and \( \vec{c} \), and orthogonal to \( \vec{a} \)

Explanation:
1. \( \vec{a}×(\vec{b}×\vec{c})=(\vec{a}·\vec{c})\vec{b}−(\vec{a}·\vec{b})\vec{c} \) is a combination of \( \vec{b} \) and \( \vec{c} \), hence coplanar with them.
2. It is a cross product with \( \vec{a} \) as a factor, so it is perpendicular to \( \vec{a} \).

Teacher's Note:
1. The triple product lies in the plane of the two vectors inside the bracket.
2. It is perpendicular to the vector outside the bracket.

Answer: (b) \( −2\vec{a} \)

Explanation:
1. \( (\vec{a}×\hat{i})×\hat{i}=−\hat{i}×(\vec{a}×\hat{i}) \).
2. The sum is therefore \( −\sum \hat{i}×(\vec{a}×\hat{i})=−2\vec{a} \).

Teacher's Note:
1. Reverse the order of the outer cross product to reuse the standard sum.
2. The sign flips.

Answer: (b) 0

Explanation:
1. \( (\vec{a}×\vec{b})·(\vec{c}×\vec{d})=(\vec{a}·\vec{c})(\vec{b}·\vec{d})−(\vec{a}·\vec{d})(\vec{b}·\vec{c}) \).
2. Directly, \( \vec{a}×\vec{b}=(−2,2,0) \) and \( \vec{c}×\vec{d}=(2,2,0) \).
3. Their dot product is \( −4+4+0=0 \), which agrees with the identity \( 1·1−1·1=0 \).

Teacher's Note:
1. Either use the four-vector identity or compute the two cross products.
2. Both give the same result.

Answer: (d) −1

Explanation:
1. The identity states \( (\vec{a}×\vec{b})·(\vec{c}×\vec{d})=(\vec{a}·\vec{c})(\vec{b}·\vec{d})−(\vec{a}·\vec{d})(\vec{b}·\vec{c}) \).
2. Comparing, \( K=−1 \).

Teacher's Note:
1. Remember the determinant form with rows \( \vec{a}·\vec{c}, \vec{a}·\vec{d} \) and \( \vec{b}·\vec{c}, \vec{b}·\vec{d} \).
2. The cross term carries the minus sign.

Answer: (a) 0

Explanation:
1. Write each term as a 2 × 2 determinant of dot products.
2. \( (\vec{b}·\vec{a})(\vec{c}·\vec{d})−(\vec{b}·\vec{d})(\vec{c}·\vec{a}) \) + \( (\vec{c}·\vec{b})(\vec{a}·\vec{d})−(\vec{c}·\vec{d})(\vec{a}·\vec{b}) \) + \( (\vec{a}·\vec{c})(\vec{b}·\vec{d})−(\vec{a}·\vec{d})(\vec{b}·\vec{c}) \).
3. All six products cancel in pairs, so the sum is 0.

Teacher's Note:
1. Expand every term with the four-vector identity.
2. Cancellation of all six terms is the whole proof.